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Solitaire II by Xanthippe

Posted by shirleycurran on 12 December 2014

DODOThere wasn’t going to be a Numpty blog this week as it can be no secret that one Numpty is hopeless with the numericals and the other one filled this one in in a rush as we were heading south to the Budock Vean crossword gathering (you can see what fun we had on Derek Harrison’s website)  via the Listener get-together that Quinapalus organised in Cambridge (where we just missed meeting Xanthippe who left minutes before we arrived!)

Aren’t friends great? Two have very generously given me their solving paths, which differ slightly from the one the other Numpty followed and intrinsically differ too. Both were full of admiration for this numerical which was an example of the pursuit of pure logic with no need for deep mathematical understanding.

My Numpty comment is that I could play the solitaire and reach a conclusion that produced three birds to complement the SOLITAIRE of the title and left the O of DODO in the centre cell, but couldn’t have created my solitaire set in the first place, so thumbs down for me – and Xanthippe didn’t give any clues that admitted him into the Oenophile Listener Setters Elite – however, he redeemed himself by leaving an empty glass on the table at the Castle in Cambridge – so all is well on that score. Many thanks Xanthippe and thanks, too, to both of the generous new LWO bloggers!

Here’s the first path by a speedy and gifted solver:

Solitaire II by Xanthippe

All answers are between 100 (256 dec) and FFE (4094); i.e. three hex digits with no leading 0’s.

I re-use x, y and ? to represent unknown digits throughout.

There are 10 properties:
P1: only numeric digits
P2: only letter digits
P3: only odd digits
P4: has repeated digit
P5:  > E00
P6: multiple of 6
P7: multiple of 7
P8: multiple of 10 (16 dec)
P9: multiple of 64 (100 dec)
P10: odd

10dn is a multiple of 7, 16 dec and 100 dec; so it is a multiple of 2800 decimal.  5600 is too big; so 10dn=2800=AF0.

By P8, 4ac, 11ac, 16ac all end in 0 and no other entries end in 0

By P5, 15dn starts with E or F.  If it starts with F, then so does the bigger 14ac.  But the 14a would be FFx and have a repeated digit (P4).  So 15dn starts with E.  14a = FEx for some x (and x must be a letter).  Whatever that x is, 16a is either xx0 or x00.  It cannot be the latter as the above shows that 9dn does not end in 0.  Also x must be even since 14a is not odd.  x cannot be F since 14ac has no repeats.  So x=A or C.  But A00 is not divisible by 6.  x must be C: 14ac = FEC, 16ac=CC0.

16dn = Cxy where x and y are both letters.  17ac has all odd digits, so x is odd.  The odd letters are B, D and F.  16dn<16ac=CC0 => x=B.  y is an even letter since 16dn is not odd and 16dn has no repeats so it is either CBA or CBE.   CBE (3262) is not divisible by 6.  So 16dn=CBA.

16dn < 9dn < 16ac.  CBA < 9dn=xyC < CC0 => 9dn=CBC

13a > 16a=CC0; 5dn is odd, so 13dn begins with an odd digit >= C.  So 13dn starts with D and 5dn = xyD.  5dn consists of all odd, letter digits. 5dn>16a, so x=D. 5dn cannot be DDD (the preamble states that no entry has all three digits the same.)  So 5dn = DBD or DFD.  But 11a<16dn=CBA so cannot begin with F.  5dn=DBD.

13ac is all letters, has no repeats and has an even digit so it is either DAF, DCF or DEF.  DCF=3535 is divisible by 7, so 13ac is either DAF or DEF.  16a=CC0 < 6dn < 5dn=DBD so 6dn starts with either C or D.  If 6dn starts with C then 5ac=DCx, but 5ac has a repeat and would then be DCC or DCD.  But it must have a numeric digit.  Hence 6dn starts with D.  11ac starts with B, ends in 0 and has a repeat so is either BB0 or B00.  If it were BB0, then 6dn would be all letters.  So 11ac=B00 and 6dn=D0A (3338) or D0E (3342).  6dn is divisible by 6, so 6dn=D0E and 13a=DEF.

8ac = xCA where x is numeric and odd (since 3dn is odd).  Of 1CA (458), 3CA (970), 5CA (1482), 7CA (1994) and 9CA (2506) only 5CA is divisible by 6.  8ac=5CA (and 8dn=58c).

11a=B00 < 3dn < 16dn=CBA.  3dn=x05, so x=B or C.  B05 (2821) is divisible by 7. So 3dn=C01.

5ac=DDx.  That x is the first digit of 7ac < 8ac=5CA.  It is also even since 1dn is not odd.  x=2 or 4.  But DD4 (3540) is divisible by 6.  So 5ac=DD2.  (That also fills in 7dn=20F).

1dn=x62 < 7dn=20F => 1dn=162.

1ac = 1xC has a repeat => 1ac = 11C or 1CC.  2dn>11a=B00 => 1ac=1CC.

2dn = Cxx < 3dn=C01 => 2dn = C01 (3073) or C03 (3075).  2dn is divisible by 7 => 2dn=C01, 7ac=215.

4ac = x00.  4ac > 8ac=5CA.  4ac must be one of 600 (1536), 700 (1792), 800 (2048), 900 (2304.  Of those only 600, 900 are divisible by 6.  12ac = xBF.  x must be an even, numeric digit since both B & F are odd letters.  12ac > 8ac = 5CA.  Hence x is 6 or 8.  4ac < 12ac so it cannot start with 9.  4ac = 600.

17ac is sandwiched between 4ac and 12ac and is all odd so it must start with 7, and 12ac must start with 8. 12a=8BF.

14dn=F7x is even, so x is even.  18a<8dn=58C.  So x=2 or 4.  F72 (3954) is divisible by 6, so 14dn=F74.

At this point the grid is completely filled except for 15dn = Exy (along with 17a and 18a).  There are twelve possibilities for 15dn, such that it along with 17ac and 18ac conform to the property restrictions.  However, note that neither 3 nor 9 appears in the grid.  Step b of the game requires that both of these digits appear in the grid.  Hence 15dn = E39 (3641) or E933 (3731).  The latter is divisible by 7 so 15dn=E39, 17a=73B and 18a=49A:

 

1 C C
6 0 0
D D 2 1 5 C A
B 0 0 8 B F
D E F E C C 0
7 3 B
4 9 A

 

Now for the game:
Step a) can only be 0 jumping 0:

1 C C
6 0 0
D D 2 1 5 C A
B 0 8 B F
D E F E C C 0
7 3 B
4 9 A

 

Step b) can be derived without too much trial and error using jumping pegs and jump direction in this order:
7 N, D in row 5 col 1 E, E in col 4 W, C in row 5 col 6 W, A in last row N, 8 S, 4 E  then N, 2 S, 1 in row 1 S, 0 in row 2 col 5 S then S then W then N then N resulting in (with I and O):

C C
O O
D D I C A
B O B F
E C O

 

Step C: Somehow (I used TEA) one determines that the three words are birds BECCAFICO, DOD and COB (SOLITAIRE is a bird as well).  BEC can be done using the D now found in col 1 (S then E then E).  Next jump A W, I E, O in col 7 N then W, C in row 1 col 5 W, then S:

O
D C O
O B
D

 

Next come COB using the D in col 2 moving E, E then S.  Using the same D pin jump W then N then jump the final O over that D to give DODO and leave that O in the center.

Here’s the second path to the creation of the solitaire set from the other brilliant solver:

Listener 4321: Solitaire II by Xanthippe 

From Solitaire game instructions, each digit 2-9 can only appear once in the grid

From clues, many cells can be identified as being either numbers or letters, odd or even, and if divisible by 16 must end in 0 (which can be entered immediately)

14ac contains only letter digits and >E00

13ac is odd and so its last digit (=first digit of 14ac) must be F(=15) rather than E(=14)

15dn>E00; its first digit (=second digit of 14ac) must be E as 14ac has no repeated digits

Since 14ac is not odd, last digit must be A or C (not E or else 14ac has repeated digit)

FEA (=4074) is a multiple of both 6 and 7

So 14ac = FEC

16ac (ending in 0) begins with C

16ac does have a repeated digit; since 9d only contains letters, must be the C

So 16ac = CC0 (=3264 which is a multiple of 6).

16dn<16ac; since 17ac is odd and ends in a letter, this must be B

So 16ac = CBA (=3258 which is a multiple of 6)

And 9dn = CBC as it is >16dn

6dn is >16ac but <E00 so must begin with D (middle digit is 0), as must 5ac/dn and 13ac

5dn (D?D) < 5ac (DD?) but 5dn contains only odd letters (B,D,F)

So 5dn = DBD, making 11ac = B00

13ac>5ac forces 13ac = DEF which makes 6dn = D0E

10dn is multiple of 7, 16 & 100; smallest possible answer is 2800 (=AF0), next 5600

5600>FFF and so 10dn = AF0

8ac is ?CA where ? is an odd number. By trial and error, 8ac = 5CA

3dn is ?05 where ? is either A or C; 3dn>10dn (=AF0) so 3dn is C05

2dn<3dn (=C05) so second digit must be 0; last digit (an odd number) must be 1 or 3

C03 (=3075) is not a multiple of 7, so 2dn = C01 (=3073 which is a multiple of 7)

4ac>8ac (=5CA) and all numbers, so first digit >5; and is a multiple of 6

By trial and error, can only be 600 or 900 but if 900 then no place in grid for a 6

So 4ac = 600

And the only place left in grid for an 8 is at start of 12ac, so 12ac = 8BF

17ac<12ac so first digit must be 7

14dn is F7? where ? is either 2 or 4

F72 is a multiple of 6 so 14dn = F74

1dn contains only numbers in format ?62 or ?64 with ? less than 5

? cannot be 4 otherwise 1dn>7dn

? cannot be 3 otherwise 1ac is 3CC (=972) which is a multiple of 6

So ? is 1 or 2 which leaves only spaces available in grid for 3 & 9 in 15dn

15dn = E39 (as E93 is a multiple of 7), making 17ac = 73B and 18ac = 49A

5ac is DD? with ? either 2 or 4 but DD4 (=3540) is a multiple of 6, so 5ac = DD2

So 7ac = 215 and 7dn = 20F

1dn<7dn so 1dn = 162 making 1ac = 1CC

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